Question: What is the average rate of change of the function $h(x)=e^{x}$ over the interval $[2,t]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^{ t-2}}{t}$ (Choice B) B $\dfrac{e^{ t}-e^{ 2}}{2}$ (Choice C) C $\dfrac{e^{ t}-e^{ 2}}{t-2}$ (Choice D) D $\dfrac{e^{ t-2}}{t-2}$
Explanation: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We are interested in the average rate of change of $h(x)=e^{x}$ over the interval $[2,t]$ : $\begin{aligned} &\phantom{=}\dfrac{h(t)-h(2)}{(t)-(2)} \\\\ =&=\dfrac{e^{t}-e^{2}}{t-2} \end{aligned}$ The average rate of change of the function is $\dfrac{e^{t}-e^{2}}{t-2}$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.